how do you calculate the moles of s2o3 2- ions consumed in each trial. Trial 2.

Question

how do you calculate the moles of s2o3 2- ions consumed in each trial. Trial 2. Solution A: DI water: 6.0 mL Buffer: 1.0 mL 0.02M Na2S2O3: 1.0 mL Starch: 5 drops 0.30M KI: 2.0mL Solution B: 0.10M H2O2: 3.0 mL Trial 2 solution A & b are mixed. The equations for both solutions are in that picture. Please help me solve this and show steps!
When Solutions A and B are mixed, the H2O2 reacts with the I (Equation 24.2). 2 I (aa) H20,(aq) 2H30 (aq)- L (aq) 4 H20() (repeat of Equation 24.2) To prevent an equilibrium (a back reaction) from occurring in Equation 24.2, thio- sulfate ion is added to the system for the purpose ofremoving I2 as it is formed: 2 S20,2 (aq) Io(aq) 21 (aq) Saoe: (aa) (24.9)

Explanation / Answer

First you need to understand that I2 is common in both the reaction. so we will start by finding mole of I2 to find out moles of S2O32- ion

in eqn 24.2 I2 is formed because of KI and peroxide

so we need to find out which of them is limiting reactant

moles of KI = 0.3*2 = 0.6 milli moles

moles of peroxide (H2O2)= 0.1 *3 = 0.3 milli moles

Now two moles of KI need 1 mole of H2O2 for reaction

which is satisfied so both are in stoichiometirc amount hence none of them is limiting reactant

moles of I2 formed need 1 mole of peroxide

so moles of I2= 0.3 millimoles

1 mole I2 need 2 moles of S2O32-

so moles of S2O32- required = 0.6 millimoles= 6*10-4 moles

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